**解题思路 **
Binary Search
The minimum element must satisfy one of two conditions: 1) If rotate, A[min] < A[min - 1]; 2) If not, A[0].
Therefore, we can use binary search:
check the middle element, if it is less than previous one, then it is minimum.
If not, there are 2 conditions as well:
If it is greater than both left and right element, then minimum element should be on its right,
otherwise if it is less than right element, then minimum element should be on its left.
** when num[mid] == num[right], we couldn't sure the position of minimum in mid's left or right, so just let right reduce one. **